A) \[20\pi \]
B) 100
C) 40p
D) \[100\pi \]
Correct Answer: C
Solution :
At centre \[{{v}_{\text{max}}}=a\omega =a.\frac{2\pi }{T}=\frac{0.2\times 2\pi }{0.01}=40\pi \]You need to login to perform this action.
You will be redirected in
3 sec